Convert to Polar Coordinates (-5,5 square root of 3)

$(- 5, 5 \sqrt{3})$

Step 1

Convert from rectangular coordinates $(x, y)$ to polar coordinates $(r, θ)$ using the conversion formulas.

$r = \sqrt{x^{2} + y^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 2

Replace $x$ and $y$ with the actual values.

$r = \sqrt{{(- 5)}^{2} + {(5 \sqrt{3})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

Simplify the expression.

Raise $- 5$ to the power of $2$ .

$r = \sqrt{25 + {(5 \sqrt{3})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Apply the product rule to $5 \sqrt{3}$ .

$r = \sqrt{25 + 5^{2} {\sqrt{3}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Raise $5$ to the power of $2$ .

$r = \sqrt{25 + 25 {\sqrt{3}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{25 + 25 {\sqrt{3}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$r = \sqrt{25 + 25 {(3^{\frac{1}{2}})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$r = \sqrt{25 + 25 \cdot 3^{\frac{1}{2} \cdot 2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Combine $\frac{1}{2}$ and $2$ .

$r = \sqrt{25 + 25 \cdot 3^{\frac{2}{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Cancel the common factor of $2$ .

Cancel the common factor.

$r = \sqrt{25 + 25 \cdot 3^{\frac{2}{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite the expression.

$r = \sqrt{25 + 25 \cdot 3}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{25 + 25 \cdot 3}$

$θ = t a n^{- 1} (\frac{y}{x})$

Evaluate the exponent.

$r = \sqrt{25 + 25 \cdot 3}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{25 + 25 \cdot 3}$

$θ = t a n^{- 1} (\frac{y}{x})$

Simplify the expression.

Multiply $25$ by $3$ .

$r = \sqrt{25 + 75}$

$θ = t a n^{- 1} (\frac{y}{x})$

Add $25$ and $75$ .

$r = \sqrt{100}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite $100$ as $10^{2}$ .

$r = \sqrt{10^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Pull terms out from under the radical, assuming positive real numbers.

$r = 10$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 10$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 10$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

Replace $x$ and $y$ with the actual values.

$r = 10$

$θ = t a n^{- 1} (\frac{5 \sqrt{3}}{- 5})$

Step 5

The inverse tangent of $- \sqrt{3}$ is $θ = 120 °$ .

$r = 10$

$θ = 120 °$

Step 6

This is the result of the conversion to polar coordinates in $(r, θ)$ form.

$(10, 120 °)$

Convert to Polar Coordinates (-5,5 square root of 3)

Related Questions

Support LCMGCF.com by disabling your adblocker.